Matching Problem 暴力

Matching Problem 暴力

题意：

给一个序列a，长度为n，序列b，长度为4，求a的子序列中，有多少个长度与b相等且当(b_i = b_j) 时 (a_i) = (a_j)

题解：

因为n<300，所以用了一个三重循环，暴力求解，注意的是，子序列不用去重。

#include <cstdio>
int cnt[310], c[310][310];
int main() {
    int n, ans = 0, a[310], b[5];
    scanf("%d", &n);
    for(int i = 0; i < n; i++) scanf("%d", &a[i]);
    for(int i = 0; i < 4; i++) scanf("%d", &b[i]);
    for(int i = n - 1; i >= 0; i--) {
        cnt[a[i]]++;
        for(int j = 0; j <= n; j++) c[i][j] = cnt[j];//c[i]这个位置到末尾j的数量
    }
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            //printf("%d
", i);
            if((b[0] == b[1] && a[i] != a[j]) || (b[0] != b[1] && a[i] == a[j])) continue;
            for(int k = j + 1; k < n; k++) {
                //printf("%d %d
", i, j);
                if(b[2] == b[1] && a[k] != a[j]) continue;
                if(b[2] == b[0] && a[k] != a[i]) continue;
                if(b[2] != b[1] && a[k] == a[j]) continue;
                if(b[2] != b[0] && a[k] == a[i]) continue;
                if(b[3] == b[0]) ans += c[k+1][a[i]];
                else if(b[3] == b[1]) ans += c[k+1][a[j]];
                else if(b[3] == b[2]) ans += c[k+1][a[k]];
                else {
                    ans += n - k - 1 - c[k+1][a[i]];
                    if (a[j]!=a[i]) ans -= c[k+1][a[j]];
                    if (a[k]!=a[i]&&a[k]!=a[j]) ans-=c[k+1][a[k]];
                }
            }
        }
    }
    printf("%d
", ans);
    return 0;
}

这个是去重了的

#include <cstdio>
#include <algorithm>
using std::min;
int cnt[310], c[310][310];
int vis1[310], vis2[310], vis3[310], vis[310];
int main() {
    int n, ans = 0, a[310], b[5], cc = 1;
    scanf("%d", &n);
    vis[n] = 0;
    for(int i = 0; i < n; i++) scanf("%d", &a[i]);
    for(int i = 0; i < 4; i++) scanf("%d", &b[i]);
    for(int i = n - 1; i >= 0; i--) {
        if(cnt[a[i]] == 0) vis[i] = vis[i+1] + 1;
        else vis[i] = vis[i+1];
        cnt[a[i]]++;
        for(int j = 0; j <= n; j++) c[i][j] = cnt[j];
    }
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            if((b[0] == b[1] && a[i] != a[j]) || (b[0] != b[1] && a[i] == a[j])) continue;
            for(int k = j + 1; k < n; k++) {
                if(b[2] == b[1] && a[k] != a[j]) continue;
                if(b[2] == b[0] && a[k] != a[i]) continue;
                if(b[2] != b[1] && a[k] == a[j]) continue;
                if(b[2] != b[0] && a[k] == a[i]) continue;
                if(vis1[a[i]] != 0 && vis1[a[i]] == vis2[a[j]] && vis2[a[j]] == vis3[a[k]]) continue;
                if(b[3] == b[0]) ans += c[k+1][a[i]];
                else if(b[3] == b[1]) ans += c[k+1][a[j]];
                else if(b[3] == b[2]) ans += c[k+1][a[k]];
                else {
                    if(b[0] != b[1] && b[1] != b[2] && b[0] != b[2]) ans += vis[k+1] - min(1,c[k+1][a[i]]) - min(1,c[k+1][a[j]]) - min(1,c[k+1][a[k]]);
                    else if(b[0] == b[1] && b[0] != b[2]) ans += vis[k+1] - min(1, c[k+1][a[i]]) - min(1, c[k+1][a[k]]);
                    else if(b[0] == b[2] && b[0] != b[1]) ans += vis[k+1] - min(1, c[k+1][a[i]]) - min(1, c[k+1][a[j]]);
                    else if(b[1] == b[2] && b[1] != b[0]) ans += vis[k+1] - min(1, c[k+1][a[i]]) - min(1, c[k+1][a[k]]);
                    else ans += vis[k+1] - min(1, c[k+1][a[i]]);
                }
                vis1[a[i]] = vis2[a[j]] = vis3[a[k]] = cc++;
                //printf("%d %d %d %d
", a[i], a[j], a[k], ans);
            }
        }
    }
    printf("%d
", ans);
    return 0;
}